Question 4 of 6: Question 4 (paper Question IV) — Relative Motion of Two Cars (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2018-May. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from PART B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — source reconstruction notes. algebraically this makes the force-along-y and moment-about-y equilibrium equations identically 0=0 for ANY combination of cable tensions, so no unique solution exists with that reading. Collar A sits on a fixed VERTICAL post (free to slide and rotate about the vertical axis, resisting lateral force + tipping moment — the standard "single collar/bearing" idealization), and that E must anchor on the SAME side as C for the six equilibrium equations to return a unique, physically sensible (all-positive-tension) solution; this reading is used below and corroborated by the clean result T_DE = 495 N exactly. (2) Question 4's source text states car B's speed as 30 m/s while the accompanying figure separately labels it 40 m/s — the printed problem statement's value (30 m/s) is used below, since the worded statement governs.
Question 4 (paper Question IV) — Relative Motion of Two Cars (Part B · Dynamics, equal value)
Given. Car A travels on a straight road at $v_A=50$ m/s, speeding up at $a_A=4\text{ m/s}^2$ (tangential only, straight path). Car B travels on a curved road at $v_B=30$ m/s, slowing at $3\text{ m/s}^2$ tangentially, with radius of curvature $\rho_B=200$ m; B's direction of travel makes 30° with A's direction.
Given data
Quantity
Value
$v_A$
50 m/s (straight road)
$a_A$ (tangential)
+4 m/s² (speeding up)
$v_B$
30 m/s
$a_B$ (tangential)
-3 m/s² (slowing down)
$\rho_B$
200 m
Angle between paths
30°
Find. The velocity and acceleration of car B relative to car A, $\mathbf{v}_{B/A}$ and $\mathbf{a}_{B/A}$.
Approach. Set up x-y axes with y along A's straight path. Resolve B's velocity and acceleration (tangential + normal, using $a_n=v_B^2/\rho_B$) into these axes, then subtract A's motion vectorially: $\mathbf{v}_{B/A}=\mathbf{v}_B-\mathbf{v}_A$ and $\mathbf{a}_{B/A}=\mathbf{a}_B-\mathbf{a}_A$.
Velocity components. $\mathbf{v}_A=(0,50)$ m/s. With B's tangent 30° from A's direction, $\mathbf{v}_B=30(\sin30^\circ,\cos30^\circ)=(15.00,25.98)$ m/s.
Relative velocity. $\mathbf{v}_{B/A}=\mathbf{v}_B-\mathbf{v}_A=(15.00,\,25.98-50)=(15.00,-24.02)$ m/s. $\boxed{|\mathbf{v}_{B/A}|=28.3\text{ m/s}}$, directed at $\arctan(15.00/24.02)=32.0^\circ$ from A's direction (toward B's side, back-and-across from A).
Normal acceleration of B. $a_{n,B}=v_B^2/\rho_B=30^2/200=4.5\text{ m/s}^2$, directed toward the centre of curvature (perpendicular to B's tangent).
Acceleration components. $\mathbf{a}_A=(0,4)$ m/s². Resolving B's tangential ($-3$ m/s² along its direction of travel) and normal (4.5 m/s², perpendicular) components into x-y: $\mathbf{a}_B=(2.40,-4.85)$ m/s².
Check — the direction of the normal-acceleration vector for car B depends on which side of its path the road curves toward; the calculation above assumes the centre of curvature lies to the right of B's direction of travel, consistent with the figure's description of a merge curving up-and-rightward into A's road. The magnitudes of $|\mathbf{v}_{B/A}|$ and the tangential contribution to $|\mathbf{a}_{B/A}|$ do not depend on this choice.
